SOLUTION: Solve. 4log₃2 - 2log₂x = 1 I attempted using the Power Property of Logarithms first, but then I got stuck because the bases of both logs were different.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve. 4log₃2 - 2log₂x = 1 I attempted using the Power Property of Logarithms first, but then I got stuck because the bases of both logs were different.      Log On


   

Question 545445: Solve.
4log₃2 - 2log₂x = 1
I attempted using the Power Property of Logarithms first, but then I got stuck because the bases of both logs were different.
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve.
4log₃2 - 2log₂x = 1
log3(2^4)-log2(x^2)=1
log3(16)-log2(x^2)=1
change to base 10
log(16)/log(3)-log(x^2)/log(2)=1
2.5237-log(x^2)/log(2)=1
2.5237-1=log(x^2)/log(2)
1.5237*log(2)=log(x^2)
.4587=log(x^2)
x^2=10^.4587=2.8754
x≈1.696
note: I tried solving without using a calculator but maybe a smarter tutor might know how.